1.
Let $J_1,...,J_n\triangleleft R$ and $P$ be a prime ideal of $R$ with
$P \supset \cap_{i=1}^n J_i$. Then $P \supset J_k$ for some $k$.
Futhermore if $P = \cap_{i=1}^n J_i$ then $P = J_k$ for some $k$.

Proof: (1) Suppose $P$ does not contain any of the $J_i$. Then
for all $i$, choose some $x_i \in J_i\setminus P$. Set $y = x_1...x_n$.
Then $y \in \cap_{i=1}^n \subset P$. But since $P$ is prime,
for some $k$ we have $x_k\in P$, a contradiction. Hence $P \supset J_i$.

Next suppose $P = \cap_{k=1}^n J_k$. Then $J_i\subset P\subset J_i$.

(2) If $n=1$ then $J\subset P_1$. This is the basis of an induction. We
shall show that for some $j = 1,...,n$, we have
$J\subset \cup_{i\ne j} P_i$, from which the result follows.

Suppose not. Then
for all $j=1,..,n$ choose $x_j \in J\setminus \cup_{i\ne j}P_i$.
Since $J \subset \cup_{i=1}^n P_i$ we must have $x_j \in P_j$ for each
$j = 1,...,n$. Set

\[
y = \sum_{j=1}^n x_1...x_{j-1}x_{j+1}...x_n
\]

Then $y \in J = \cup_{i=1}^n P_i$ so $y \in P_k$ for some $k =1,...,n$.
But this means